A simple random sample with n= 53 provided a sample mean of 27.5 and a sample standard deviation of 4.4. (Round your answers to one decimal place.) (a) Develop a 90% confidence interval for the population mean. to (b) Develop a 95% confidence interval for the population mean. to (c) Develop a 99% confidence interval for the population mean. to (d) What happens to the margin of error and the confidence interval as the confidence level is increased? As the confidence level increases, there is a smaller margin of error and a more narrow confidence interval. As the confidence level increases, there is a larger margin of error and a wider confidence interval. As the confidence level increases, there is a larger margin of error and a more narrow confidence interval. As the confidence level increases, there is a smaller margin of error and a wider confidence interval. A & cricut
A simple random sample with n= 53 provided a sample mean of 27.5 and a sample standard deviation of 4.4. (Round your answers to one decimal place.) (a) Develop a 90% confidence interval for the population mean. to (b) Develop a 95% confidence interval for the population mean. to (c) Develop a 99% confidence interval for the population mean. to (d) What happens to the margin of error and the confidence interval as the confidence level is increased? As the confidence level increases, there is a smaller margin of error and a more narrow confidence interval. As the confidence level increases, there is a larger margin of error and a wider confidence interval. As the confidence level increases, there is a larger margin of error and a more narrow confidence interval. As the confidence level increases, there is a smaller margin of error and a wider confidence interval. A & cricut
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![To tackle the problem provided in the exercise, you need to determine confidence intervals for a population mean using given sample statistics. The example in the image involves calculating 90%, 95%, and 99% confidence intervals based on data from a random sample.
Here is the given problem transcribed for educational purposes:
---
### Confidence Interval Calculation Exercise
You may need to use the appropriate appendix table or technology to answer this question.
Given:
- A simple random sample with \( n = 53 \) provided a sample mean of 27.5 and a sample standard deviation of 4.4. (Round your answers to one decimal place).
#### (a) Develop a 90% confidence interval for the population mean.
\[ \text{Interval: } \_\_\_\_ \text{ to } \_\_\_\_ \]
#### (b) Develop a 95% confidence interval for the population mean.
\[ \text{Interval: } \_\_\_\_ \text{ to } \_\_\_\_ \]
#### (c) Develop a 99% confidence interval for the population mean.
\[ \text{Interval: } \_\_\_\_ \text{ to } \_\_\_\_ \]
#### (d) What happens to the margin of error and the confidence interval as the confidence level is increased?
- O As the confidence level increases, there is a smaller margin of error and a more narrow confidence interval.
- O As the confidence level increases, there is a larger margin of error and a wider confidence interval.
- O As the confidence level increases, there is a larger margin of error and a more narrow confidence interval.
- O As the confidence level increases, there is a smaller margin of error and a wider confidence interval.
---
To find the confidence intervals, you will use the formula for the confidence interval for the population mean:
\[
\text{Confidence Interval} = \bar{x} \pm z^* \left( \frac{\sigma}{\sqrt{n}} \right)
\]
where:
- \( \bar{x} \) is the sample mean (27.5 in this case),
- \( z^* \) is the critical value (depends on the confidence level),
- \( \sigma \) is the sample standard deviation (4.4 in this case),
- \( n \) is the sample size (53 in this case).
The corresponding \( z^* \) values for common confidence](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa4ed8498-f2f2-4dcb-bcb0-3b3cd025dd21%2F0bec2e72-cc0f-4aac-b12d-8c395f432d7b%2Ff1k8efd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:To tackle the problem provided in the exercise, you need to determine confidence intervals for a population mean using given sample statistics. The example in the image involves calculating 90%, 95%, and 99% confidence intervals based on data from a random sample.
Here is the given problem transcribed for educational purposes:
---
### Confidence Interval Calculation Exercise
You may need to use the appropriate appendix table or technology to answer this question.
Given:
- A simple random sample with \( n = 53 \) provided a sample mean of 27.5 and a sample standard deviation of 4.4. (Round your answers to one decimal place).
#### (a) Develop a 90% confidence interval for the population mean.
\[ \text{Interval: } \_\_\_\_ \text{ to } \_\_\_\_ \]
#### (b) Develop a 95% confidence interval for the population mean.
\[ \text{Interval: } \_\_\_\_ \text{ to } \_\_\_\_ \]
#### (c) Develop a 99% confidence interval for the population mean.
\[ \text{Interval: } \_\_\_\_ \text{ to } \_\_\_\_ \]
#### (d) What happens to the margin of error and the confidence interval as the confidence level is increased?
- O As the confidence level increases, there is a smaller margin of error and a more narrow confidence interval.
- O As the confidence level increases, there is a larger margin of error and a wider confidence interval.
- O As the confidence level increases, there is a larger margin of error and a more narrow confidence interval.
- O As the confidence level increases, there is a smaller margin of error and a wider confidence interval.
---
To find the confidence intervals, you will use the formula for the confidence interval for the population mean:
\[
\text{Confidence Interval} = \bar{x} \pm z^* \left( \frac{\sigma}{\sqrt{n}} \right)
\]
where:
- \( \bar{x} \) is the sample mean (27.5 in this case),
- \( z^* \) is the critical value (depends on the confidence level),
- \( \sigma \) is the sample standard deviation (4.4 in this case),
- \( n \) is the sample size (53 in this case).
The corresponding \( z^* \) values for common confidence
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